Is double-braced scalar initialization allowed by the C++ standard?

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I have the following code:

int x = ;

Is this syntax valid according to the C++ standard? (I'm interested in C++11 and later.)

When using the latest compilers there is no problem, however in some older ones (e.g. GCC 4.8.5) it gives following error:

error: braces around scalar initializer for type 'int'

c++ c++11 language-lawyer

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edited Aug 29 at 13:00

Boann

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asked Aug 29 at 10:34

Igor

35939

add a comment | 

up vote
31
down vote

favorite

2

I have the following code:

int x = ;

Is this syntax valid according to the C++ standard? (I'm interested in C++11 and later.)

When using the latest compilers there is no problem, however in some older ones (e.g. GCC 4.8.5) it gives following error:

error: braces around scalar initializer for type 'int'

c++ c++11 language-lawyer

share|improve this question

edited Aug 29 at 13:00

Boann

35.6k1184116

asked Aug 29 at 10:34

Igor

35939

add a comment | 

up vote
31
down vote

favorite

2

up vote
31
down vote

favorite

2

2

I have the following code:

int x = ;

Is this syntax valid according to the C++ standard? (I'm interested in C++11 and later.)

When using the latest compilers there is no problem, however in some older ones (e.g. GCC 4.8.5) it gives following error:

error: braces around scalar initializer for type 'int'

c++ c++11 language-lawyer

share|improve this question

edited Aug 29 at 13:00

Boann

35.6k1184116

asked Aug 29 at 10:34

Igor

35939

I have the following code:

int x = ;

Is this syntax valid according to the C++ standard? (I'm interested in C++11 and later.)

When using the latest compilers there is no problem, however in some older ones (e.g. GCC 4.8.5) it gives following error:

error: braces around scalar initializer for type 'int'

c++ c++11 language-lawyer

share|improve this question

edited Aug 29 at 13:00

Boann

35.6k1184116

asked Aug 29 at 10:34

Igor

35939

share|improve this question

share|improve this question

edited Aug 29 at 13:00

Boann

35.6k1184116

edited Aug 29 at 13:00

Boann

35.6k1184116

edited Aug 29 at 13:00

Boann

35.6k1184116

35.6k1184116

asked Aug 29 at 10:34

Igor

35939

asked Aug 29 at 10:34

Igor

35939

asked Aug 29 at 10:34

Igor

35939

35939

add a comment | 

add a comment | 

1 Answer
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up vote
33
down vote



accepted

This is ill-formed. gcc is wrong to accept it and clang seems to allow it as an extension, as it warns about it.

I'm going to quote the latest draft, but it doesn't make a difference. List initialization works as follows as per [dcl.init.list], where T is int in this case:

• If the initializer list is a designated initializer list, [...] => it's not


• If T is an aggregate class [...] => it's not


• If T is a character array [...] => it's not.


• If T is an aggregate [...] => it's not (only arrays and classes are aggregates)


• If the initializer list has no elements [...] => it doesn't


• If T is a specialization of std::initializer_list [...] => it's not


• If T is a class type [...] => it's not


• If T is an enumeration with fixed underlying type [...] => it's not


• If the initializer list has a single element of type E [...] => a braced initializer list has no type, so no


• If T is a reference type [...] => it isn't


• If the initializer list has no elements [...] => it doesn't


• Otherwise the program is ill-formed

share|improve this answer

answered Aug 29 at 10:48

Rakete1111

31.9k973107

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  +1 and see also core language issue 1501 which follows the same logic as this answer to explicitly say that it's invalid.
  – hvd
  Aug 29 at 11:51
  

  
  
  
  

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

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active

oldest

votes

up vote
33
down vote



accepted

This is ill-formed. gcc is wrong to accept it and clang seems to allow it as an extension, as it warns about it.

I'm going to quote the latest draft, but it doesn't make a difference. List initialization works as follows as per [dcl.init.list], where T is int in this case:

• If the initializer list is a designated initializer list, [...] => it's not


• If T is an aggregate class [...] => it's not


• If T is a character array [...] => it's not.


• If T is an aggregate [...] => it's not (only arrays and classes are aggregates)


• If the initializer list has no elements [...] => it doesn't


• If T is a specialization of std::initializer_list [...] => it's not


• If T is a class type [...] => it's not


• If T is an enumeration with fixed underlying type [...] => it's not


• If the initializer list has a single element of type E [...] => a braced initializer list has no type, so no


• If T is a reference type [...] => it isn't


• If the initializer list has no elements [...] => it doesn't


• Otherwise the program is ill-formed

share|improve this answer

answered Aug 29 at 10:48

Rakete1111

31.9k973107

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  +1 and see also core language issue 1501 which follows the same logic as this answer to explicitly say that it's invalid.
  – hvd
  Aug 29 at 11:51
  

  
  
  
  

add a comment | 

up vote
33
down vote



accepted

This is ill-formed. gcc is wrong to accept it and clang seems to allow it as an extension, as it warns about it.

I'm going to quote the latest draft, but it doesn't make a difference. List initialization works as follows as per [dcl.init.list], where T is int in this case:

• If the initializer list is a designated initializer list, [...] => it's not


• If T is an aggregate class [...] => it's not


• If T is a character array [...] => it's not.


• If T is an aggregate [...] => it's not (only arrays and classes are aggregates)


• If the initializer list has no elements [...] => it doesn't


• If T is a specialization of std::initializer_list [...] => it's not


• If T is a class type [...] => it's not


• If T is an enumeration with fixed underlying type [...] => it's not


• If the initializer list has a single element of type E [...] => a braced initializer list has no type, so no


• If T is a reference type [...] => it isn't


• If the initializer list has no elements [...] => it doesn't


• Otherwise the program is ill-formed

share|improve this answer

answered Aug 29 at 10:48

Rakete1111

31.9k973107

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  +1 and see also core language issue 1501 which follows the same logic as this answer to explicitly say that it's invalid.
  – hvd
  Aug 29 at 11:51
  

  
  
  
  

add a comment | 

up vote
33
down vote



accepted

up vote
33
down vote



accepted

This is ill-formed. gcc is wrong to accept it and clang seems to allow it as an extension, as it warns about it.

I'm going to quote the latest draft, but it doesn't make a difference. List initialization works as follows as per [dcl.init.list], where T is int in this case:

• If the initializer list is a designated initializer list, [...] => it's not


• If T is an aggregate class [...] => it's not


• If T is a character array [...] => it's not.


• If T is an aggregate [...] => it's not (only arrays and classes are aggregates)


• If the initializer list has no elements [...] => it doesn't


• If T is a specialization of std::initializer_list [...] => it's not


• If T is a class type [...] => it's not


• If T is an enumeration with fixed underlying type [...] => it's not


• If the initializer list has a single element of type E [...] => a braced initializer list has no type, so no


• If T is a reference type [...] => it isn't


• If the initializer list has no elements [...] => it doesn't


• Otherwise the program is ill-formed

share|improve this answer

answered Aug 29 at 10:48

Rakete1111

31.9k973107

This is ill-formed. gcc is wrong to accept it and clang seems to allow it as an extension, as it warns about it.

I'm going to quote the latest draft, but it doesn't make a difference. List initialization works as follows as per [dcl.init.list], where T is int in this case:

• If the initializer list is a designated initializer list, [...] => it's not


• If T is an aggregate class [...] => it's not


• If T is a character array [...] => it's not.


• If T is an aggregate [...] => it's not (only arrays and classes are aggregates)


• If the initializer list has no elements [...] => it doesn't


• If T is a specialization of std::initializer_list [...] => it's not


• If T is a class type [...] => it's not


• If T is an enumeration with fixed underlying type [...] => it's not


• If the initializer list has a single element of type E [...] => a braced initializer list has no type, so no


• If T is a reference type [...] => it isn't


• If the initializer list has no elements [...] => it doesn't


• Otherwise the program is ill-formed

share|improve this answer

answered Aug 29 at 10:48

Rakete1111

31.9k973107

share|improve this answer

share|improve this answer

answered Aug 29 at 10:48

Rakete1111

31.9k973107

answered Aug 29 at 10:48

Rakete1111

31.9k973107

answered Aug 29 at 10:48

Rakete1111

31.9k973107

31.9k973107

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  +1 and see also core language issue 1501 which follows the same logic as this answer to explicitly say that it's invalid.
  – hvd
  Aug 29 at 11:51
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  +1 and see also core language issue 1501 which follows the same logic as this answer to explicitly say that it's invalid.
  – hvd
  Aug 29 at 11:51
  

  
  
  
  

4

4

+1 and see also core language issue 1501 which follows the same logic as this answer to explicitly say that it's invalid.
– hvd
Aug 29 at 11:51

+1 and see also core language issue 1501 which follows the same logic as this answer to explicitly say that it's invalid.
– hvd
Aug 29 at 11:51

add a comment | 

 

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